The empirical formula of metal iodide : CoI₃(Cobalt(III) Iodide)
<h3>Further explanation</h3>13.02 g sample of Cobalt , then mol Co(MW=58.933 g/mol) :
Mass of metal iodide formed : 97.12 g, so mass of Iodine :
Then mol iodine (MW=126.9045 g/mol) :
mol ratio of Cobalt and Iodine in the compound :
Answer:
The reaction would shift toward the reactants
When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm
Explanation:
For the reaction:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Where K is defined as:
As initial pressures of all 3 gases is 1.0atm, reaction quotient, Q, is:
As Q > K, <em>the reaction will produce more NH₃ until Q = K consuming N₂ and H₂.</em>
Thus, there are true:
<h3>The reaction would shift toward the reactants</h3><h3>When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm</h3><em />